So I understand Big-O, Big-Omega and Big-Theta conceptually, but I'm not sure how to prove Big-Omega and Big-Theta.
function f is in Big-O(g) if and only if there exists some constant c > 0 and some constant n_0 ≥ 1 such that for all n ≥ n_0, the expression f(n) ≤ c·g(n) is true.
Big-Omega is the opposite, c·g(n) ≤ f(n).
Big-Theta sandwiches c1·f(n) ≤ g(n) ≤ c2·f(n).
I need to prove/disprove if (2^{n})^{1/3} ∈ Θ(2^{n}) by using all three notations.
What I have so far:
Big-O : (2^{n})^{1/3} ≤ c·2^{n} when c=1 and n_0 = 1, so (2^{n})^{1/3} ∈ O(2^{n})
Big-Omega : We can rewrite (2^{n})^{1/3} = (1/(2^{2n/3}))·(2^n). We see that for c·g(n) ≤ f(n), c has to be ≤ 1/(2^{2n/3}) which is not possible since c > 0. So, there does not exist a c > 0 that satisfies c·g(n) ≤ f(n) and thus, (2^{n})^{1/3} ∉ Ω(2^{n})
Big-Theta : Since (2^{n})^{1/3} ∉ Ω(2^{n}), there is no lower bound c1·f(n) ≤ g(n). Therefore, (2^{n})^{1/3} ∉ Θ(2^{n})
Is this how you are supposed to prove it?
First simplify f(n) = (2^n)^(1/3) to f(n) = 2^(n/3). Then, take a limit of lim_{n\to\infty} f(n)/g(n) that g(n) = 2^n:
lim_{n\to\infty} 2^(n/3) / 2^n = lim_{n\to\infty} 1 / 2^(2n/3) = 0
Hence, f(n) = o(g(n)) (little-oh). It means f(n) is not in \Theta(g(n)). Notice that f(n) = O(g(n)) (Big-Oh) as it is in o(g(n)) (little-Oh).
Related
c f(n) is in theta f(n) for c>0
I know that c is a constant, and if I can prove c f(n) is in big O(f(n) and in big Omega f(n) simultaneously, it is also in theta f(n), but how can I prove? I got confused.
c f(n) is O(f(n)) because there is a constant k such that :
|c f(n)| ≤ k |f(n)| as n -> infinity
Hence, |c| |f(n)| ≤ k |f(n)|
dividing both sides by |f(n)| we get |c| ≤ k
So, any value of k larger than |c| would satisfy this condition. Therefore, c f(n) is O(f(n))
You can use the same method to show that c f(n) is also Ω(f(n)), and therefore it is ϴ(f(n))
I have been given the problem:
f(n) are asymptotically positive functions. Prove f(n) = Θ(g(n)) iff g(n) = Θ(f(n)).
Everything I have found points to this statement being invalid. For example an answer I've come across states:
f(n) = O(g(n)) implies g(n) = O(f(n))
f(n) = O(g(n)) means g(n) grows faster than f(n). It cannot imply that f(n) grows
faster than g(n). Hence not true.
Another states:
If f(n) = O(g(n)) then O(f(n)). This is false. If f(n) = 1 and g(n) = n
for all natural numbers n, then f(n) <= g(n) for all natural numbers n, so
f(n) = O(g(n)). However, suppose g(n) = O(f(n)). Then there are natural
numbers n0 and a constant c > 0 such that n=g(n) <= cf(n) = c for all n >=
n0 which is impossible.
I understand that there are slight differences between my exact question and the examples I have found, but I've only been able to come up with solutions that do not prove it. I am correct in thinking that it is not able to be proved or am I looking over some detail?
You can start from here:
Formal Definition: f(n) = Θ (g(n)) means there are positive constants c1, c2, and k, such that 0 ≤ c1g(n) ≤ f(n) ≤ c2g(n) for all n ≥ k.
Because you have that iff, you need to start from the left side and to prove the right side, and then start from the right side and prove the left side.
Left -> right
We consider that:
f(n) = Θ(g(n))
and we want to prove that
g(n) = Θ(f(n))
So, we have some positive constants c1, c2 and k such that:
0 ≤ c1*g(n) ≤ f(n) ≤ c2*g(n), for all n ≥ k
The first relation between f and g is:
c1*g(n) ≤ f(n) => g(n) ≤ 1/c1*f(n) (1)
The second relation between f and g is:
f(n) ≤ c2*g(n) => 1/c2*f(n) ≤ g(n) (2)
If we combine (1) and (2), we obtain:
1/c2*f(n) ≤ g(n) ≤ 1/c1*f(n)
If you consider c3 = 1/c2 and c4 = 1/c1, they exist and are positive (because the denominators are positive). And this is true for all n ≥ k (where k can be the same).
So, we have some positive constants c3, c4, k such that:
c3*f(n) ≤ g(n) ≤ c4*f(n), for all n ≥ k
which means that g(n) = Θ(f(n)).
Analogous for right -> left.
So I'm struggling with proving (or disproving) the above question. I feel like it is true, but I'm not sure how to show it.
Again, the question is if g(n) is o(f(n)), then f(n) + g(n) is Theta(f(n))
Note, that is a little-o, not a big-o!!!
So far, I've managed to (easily) show that:
g(n) = o(f(n)) -> g(n) < c*f(n)
Then g(n) + f(n) < (c+1)*f(n) -> (g(n) + f(n)) = O(f(n))
However, for showing Big Omega, I'm not sure what to do there.
Am I going about this right?
EDIT: Everyone provided great help, but I could only mark one. THANK YOU.
One option would be to take the limit of (f(n) + g(n)) / f(n) as n tends toward infinity. If this converges to a finite, nonzero value, then f(n) + g(n) = Θ(f(n)).
Assuming that f(n) is nonzero for sufficiently large n, the above ratio, in the limit, is
(f(n) + g(n)) / f(n)
= f(n) / f(n) + g(n) / f(n)
= 1 + g(n) / f(n).
Therefore, taking the limit as n goes to infinity, the above expression converges to 1 because the ratio goes to zero (this is what it means for g(n) to be o(f(n)).
So far so good.
For the next step, recall that in the best case, 0 <= g(n); this should get you a lower bound on g(n) + f(n).
Before we begin, lets first state what little-o and Big-Theta notations means:
Little-o notation
Formally, that g(n) = o(f(n)) (or g(n) ∈ o(f(n))) holds for
sufficiently large n means that for every positive constant ε
there exists a constant N such that
|g(n)| ≤ ε*|f(n)|, for all n > N (+)
From https://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation.
Big-Θ notation
h(n) = Θ(f(n)) means there exists positive constants k_1, k_2
and N, such that k_1 · |f(n)| and k_2 · |f(n)| is an upper bound
and lower bound on on |h(n)|, respectively, for n > N, i.e.
k_1 · |f(n)| ≤ |h(n)| ≤ k_2 · |f(n)|, for all n > N (++)
From https://www.khanacademy.org/computing/computer-science/algorithms/asymptotic-notation/a/big-big-theta-notation.
Given: g(n) ∈ o(f(n))
Hence, in our case, for every ε>0 we can find some constant N such that (+), for our functions g(n) and f(n). Hence, for n>N, we have
|g(n)| ≤ ε*|f(n)|, for some ε>0, for all n>N
Choose a constant ε < 1 (recall, the above holds for all ε > 0),
with accompanied constant N.
Then the following holds for all n>N
ε(|g(n)| + |f(n)|) ≤ 2|f(n)| ≤ 2(|g(n)| + |f(n)|) ≤ 4*|f(n)| (*)
Stripping away the left-most inequality in (*) and dividing by 2, we have:
|f(n)| ≤ |g(n)| + |f(n)| ≤ 2*|f(n)|, n>N (**)
We see that this is the very definition Big-Θ notation, as presented in (++), with constants k_1 = 1, k_2 = 2 and h(n) = g(n)+f(n). Hence
(**) => g(n) + f(n) is in Θ(f(n))
Ans we have shown that g(n) ∈ o(f(n)) implies (g(n) + f(n)) ∈ Θ(f(n)).
Curious O(n^n) is pretty high but O(n^n^n) is higher and O(n^n^n^n) is higher still. Is there a highest big O?
Suppose there exists a maximal big O complexity given by f, that is,
g(n) ∈ O(f(n)), ∀ g:ℝ->ℝ
Then, let g(n) = f(n)^2.
Since lim f(n) = ∞,
f(n) f(n) 1
lim ──── = ────── = ──── = 0
g(n) f(n)^2 f(n)
That means g(n) ∉ O(f(n)). Contradiction.
I have these three questions for an exam review:
If f(n) = 2n - 3 give two different functions g(n) and h(n) (so g(n) doesn't equal h(n)) such that f(n) = O(g(n)) and f(n) = O(h(n))
Now do the same again with functions g'(n) and h'(n), but this time the function should be of the form
g'(n) = Ɵ(f(n)) and f(n) = o(h'(n))
Is it possible for a function f(n) = O(g(n)) and f(n) = Ω(g(n))?
I know that a function is O(n) of another, if it is less than or equal to the other function. So I think 1. could be g(n) = 2n²-3 and h(n) = 2n²-10.
I also know that a function is Ɵ(n) of another if it is basically equal to the other function (we can ignore constants), and o(n) if it is only less than the function, so for 2. I think you could have g'(n) = 2n-15 and h'(n) = 2n.
To 3.: It is possible for a function to be both O(n) and Ω(n) because O(n) and Ω(n) allows for the function to be the same as the given function, so you could have a function g(n) that equals f(n) and satisfies the rules for being both O and Ω.
Can someone please tell me if this is correct?
Your answers are mostly right. But I would like to add some points:
Given is f(n) = 2n - 3
With g(n) = 2n²-3 and h(n) = 2n²-10 f(n) is in O(g(n)) and in O(h(n)). But your g(n) and h(n) are basicly the same, at least they are both in Θ(n²). There exists many other function that would also work. E.g.
f(n) ∈ O(n) ⇒ g(n) = n
f(n) ∈ O(nk) ⇒ g(n) = nk ∀ k ≥ 1
f(n) ∈ O(2ⁿ) ⇒ g(n) = 2ⁿ
g'(n) = 2n-15 reduces to g'(n) = n, if we think in complexities, and this is right. In fact, it is the only possible answer.
But f(n) ∈ o(h'(n)) does not hold for h'(n) = 2n. Little-o means that
limn → ∞ | f(n)/g(n) | = 0 ⇔ f(n) ∈ o(g(n))
So you can choose h'(n) = n² or more general h'(n) = nk ∀ k > 1 or h'(n) = cⁿ for a constant c > 1.
Yes it is possible and you can take it also as a definition for Θ(g(n)):
f(n) ∈ Θ(g(n)) ⇔ f(n) ∈ O(g(n)) and f(n) ∈ Ω(g(n))